虽然我喜欢Martijn的回答，但像george一样，我想知道是否可以通过使用运行总和而不是在大多数相同数字上反复应用sum()来加快速度。
此外，在斜坡上升阶段默认使用None值的想法很有趣。实际上，可能会有许多不同的情况可以构思移动平均值。让我们将平均数的计算分成三个阶段：
- 斜坡上升：当前迭代次数<窗口大小的起始迭代
- 稳步进展：我们有正好窗口大小的元素可用于计算正常的
average := sum(x[iteration_counter-window_size:iteration_counter])/window_size
- 斜坡下降：在输入数据的末尾，我们可以返回另一个
window_size - 1个“平均”数字。
这是一个接受以下内容的函数：
- 任意可迭代对象（生成器也可以）作为数据输入
- 任意窗口大小>=1
- 在斜坡上升/下降期间打开/关闭值的生产的参数
- 回调函数用于控制如何生成值。这可以用于不断提供默认值（例如
None）或提供部分平均值
以下是代码：
from collections import deque
def moving_averages(data, size, rampUp=True, rampDown=True):
"""Slide a window of <size> elements over <data> to calc an average
First and last <size-1> iterations when window is not yet completely
filled with data, or the window empties due to exhausted <data>, the
average is computed with just the available data (but still divided
by <size>).
Set rampUp/rampDown to False in order to not provide any values during
those start and end <size-1> iterations.
Set rampUp/rampDown to functions to provide arbitrary partial average
numbers during those phases. The callback will get the currently
available input data in a deque. Do not modify that data.
"""
d = deque()
running_sum = 0.0
data = iter(data)
for count in range(1, size):
try:
val = next(data)
except StopIteration:
break
running_sum += val
d.append(val)
if rampUp:
if callable(rampUp):
yield rampUp(d)
else:
yield running_sum / size
exhausted_early = True
for val in data:
exhausted_early = False
running_sum += val
yield running_sum / size
d.append(val)
running_sum -= d.popleft()
if rampDown:
if exhausted_early:
running_sum -= d.popleft()
for (count) in range(min(len(d), size-1), 0, -1):
if callable(rampDown):
yield rampDown(d)
else:
yield running_sum / size
running_sum -= d.popleft()
这个版本似乎比Martijn的版本快一点 - 虽然Martijn的版本更加优雅。这是测试代码:
print("")
print("Timeit")
print("-" * 80)
from itertools import islice
def window(seq, n=2):
"Returns a sliding window (of width n) over data from the iterable"
" s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ... "
it = iter(seq)
result = tuple(islice(it, n))
if len(result) == n:
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
def moving_averages_SO(values, size):
for selection in window(values, size):
yield sum(selection) / size
import timeit
problems = [int(i) for i in (10, 100, 1000, 10000, 1e5, 1e6, 1e7)]
for problem_size in problems:
print("{:12s}".format(str(problem_size)), end="")
so = timeit.repeat("list(moving_averages_SO(range("+str(problem_size)+"), 5))", number=1*max(problems)//problem_size,
setup="from __main__ import moving_averages_SO")
print("{:12.3f} ".format(min(so)), end="")
my = timeit.repeat("list(moving_averages(range("+str(problem_size)+"), 5, False, False))", number=1*max(problems)//problem_size,
setup="from __main__ import moving_averages")
print("{:12.3f} ".format(min(my)), end="")
print("")
输出结果：
Timeit
--------------------------------------------------------------------------------
10 7.242 7.656
100 5.816 5.500
1000 5.787 5.244
10000 5.782 5.180
100000 5.746 5.137
1000000 5.745 5.198
10000000 5.764 5.186
现在可以通过这个函数调用来解决原始问题：
print(list(moving_averages(range(1,11), 5,
rampUp=lambda _: None,
rampDown=False)))
输出结果：
[None, None, None, None, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0]
